The RD Sharma Class 9 solutions for the chapter “Circle” exercise 16.5 is given here. In this exercise, students will learn about cyclic quadrilateral. A quadrilateral is a cyclic quadrilateral if its all vertices lie on a circle. This exercise includes several questions to help the students learn more about circles. Students are advised to click on the link below and get pdf to practice more on topic – circle. RD Sharma Solutions class 9 chapter 16 to help students in their endeavor to have a better understanding of circles.

## Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.5

### Access Answers to Maths RD Sharma Class 9 Chapter 16 Circles Exercise 16.5 Page number 16.83

**Question 1: In figure, ΔABC is an equilateral triangle. Find m∠BEC.**

**Solution:**

ΔABC is an equilateral triangle. (Given)

Each angle of an equilateral triangle is 60 degrees.

In quadrilateral ABEC:

∠BAC + ∠BEC = 180^{o} (Opposite angles of quadrilateral)

60^{o} + ∠BEC = 180^{ o}

∠BEC = 180^{ o} – 60^{ o}

∠BEC = 120^{ o}

**Question 2: In figure, Δ PQR is an isosceles triangle with PQ = PR and m∠PQR=35°. Find m∠QSR and m∠QTR.**

**Solution:**

Given: ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°

In ΔPQR:

∠PQR = ∠PRQ = 35^{o} (Angle opposite to equal sides)

Again, by angle sum property

∠P + ∠Q + ∠R = 180^{ o}

∠P + 35^{ o} + 35^{ o} = 180^{ o}

∠P + 70^{ o} = 180^{ o}

∠P = 180^{ o} – 70^{ o}

∠P = 110^{ o}

Now, in quadrilateral SQTR,

∠QSR + ∠QTR = 180^{ o} (Opposite angles of quadrilateral)

110^{ o} + ∠QTR = 180^{ o}

∠QTR = 70^{ o}

** Question 3: In figure, O is the centre of the circle. If ∠BOD = 160 ^{o}, find the values of x and y.**

**Solution: **

From figure: ∠BOD = 160^{ o}

By degree measure theorem: ∠BOD = 2 ∠BCD

160** ^{ o}** = 2x

or x = 80^{ o}

Now, in quadrilateral ABCD,

∠BAD + ∠BCD = 180** ^{ o}** (Opposite angles of Cyclic quadrilateral)

y + x = 180^{ o}

Putting value of x,

y + 80** ^{ o}** = 180

^{ o}y = 100^{ o}

Answer: x = 80** ^{ o} **and y = 100

**.**

^{ o}**Question 4: In figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100 ^{o} and ∠ABD = 70^{o}, find ∠ADB.**

**Solution: **

From figure:

In quadrilateral ABCD,

∠DCB + ∠BAD = 180** ^{o}** (Opposite angles of Cyclic quadrilateral)

100** ^{ o}** + ∠BAD = 180

^{o}∠BAD = 80^{0}

In Δ BAD:

By angle sum property: ∠ADB + ∠DAB + ∠ABD = 180^{ o}

∠ADB + 80^{o} + 70^{ o} = 180^{ o}

∠ADB = 30^{o}

**Question 5: If ABCD is a cyclic quadrilateral in which AD||BC (figure). Prove that ∠B = ∠C.**

**Solution:**

Given: ABCD is a cyclic quadrilateral with AD ‖ BC

=> ∠A + ∠C = 180^{o} ………(1)

and ∠A + ∠B = 180^{o} ………(2)

Form (1) and (2), we have

∠B = ∠C

Hence proved.

**Question 6: In figure, O is the centre of the circle. Find ∠CBD.**

**Solution: **

Given: ∠BOC = 100^{o}

By degree measure theorem: ∠AOC = 2 ∠APC

100^{ o} = 2 ∠APC

or ∠APC = 50^{ o}

Again,

∠APC + ∠ABC = 180^{ o} (Opposite angles of a cyclic quadrilateral)

50^{o} + ∠ABC = 180^{ o}

or ∠ABC = 130^{ o}

Now, ∠ABC + ∠CBD = 180^{ o} (Linear pair)

130^{o} + ∠CBD = 180^{ o}

or ∠CBD = 50^{ o}

**Question 7: In figure, AB and CD are diameters of a circle with centre O. If ∠OBD = 50 ^{0}, find ∠AOC.**

**Solution: **

Given: ∠OBD = 50^{0}

Here, AB and CD are the diameters of the circles with centre O.

∠DBC = 90** ^{0}** ….(i)

Also, ∠DBC = 50** ^{0}** + ∠OBC

90** ^{0}** = 50

**+ ∠OBC**

^{0}or ∠OBC = 40^{0}

Again, By degree measure theorem: ∠AOC = 2 ∠ABC

∠AOC = 2∠OBC = 2 x 40** ^{0}** = 80

^{0}**Question 8: On a semi-circle with AB as diameter, a point C is taken, so that m(∠CAB) = 30 ^{0}. Find m(∠ACB) and m(∠ABC).**

**Solution: **

Given: m(∠CAB)= 30^{0}

To Find: m(∠ACB) and m(∠ABC).

Now,

∠ACB = 90** ^{0}** (Angle in semi-circle)

Now,

In △ABC, by angle sum property: ∠CAB + ∠ACB + ∠ABC = 180^{0}

30** ^{0}** + 90

**+ ∠ABC = 180**

^{0}

^{0}∠ABC = 60^{0}

Answer: ∠ACB = 90** ^{0}** and ∠ABC = 60

^{0}**Question 9: In a cyclic quadrilateral ABCD if AB||CD and B = 70 ^{o} , find the remaining angles.**

**Solution: **

A cyclic quadrilateral ABCD with AB||CD and B = 70^{o}.

∠B + ∠C = 180^{o} (Co-interior angle)

70^{0} + ∠C = 180^{0}

∠C = 110^{0}

And,

=> ∠B + ∠D = 180^{0} (Opposite angles of Cyclic quadrilateral)

70^{0} + ∠D = 180^{0}

∠D = 110^{0}

Again, ∠A + ∠C = 180^{0} (Opposite angles of cyclic quadrilateral)

∠A + 110^{0} = 180^{0}

∠A = 70^{0}

Answer: ∠A = 70^{0} , ∠C = 110^{0 }and ∠D = 110^{0}

**Question 10: In a cyclic quadrilateral ABCD, if m ∠A = 3(m∠C). Find m ∠A.
Solution: **

∠A + ∠C = 180^{o} …..(1)

Since m ∠A = 3(m∠C) (given)

=> ∠A = 3∠C …(2)

Equation (1) => 3∠C + ∠C = 180^{ o}

or 4∠C = 180^{o}

or ∠C = 45^{o}

From equation (2)

∠A = 3 x 45^{o} = 135^{o}

**Question 11: In figure, O is the centre of the circle ∠DAB = 50°. Calculate the values of x and y.**

**Solution: **

Given : ∠DAB = 50^{o}

By degree measure theorem: ∠BOD = 2 ∠BAD

so, x = 2( 50^{0}) = 100^{0}

Since, ABCD is a cyclic quadrilateral, we have

∠A + ∠C = 180^{0}

50^{0} + y = 180^{0}

y = 130^{0}

## RD Sharma Solutions for Class 9 Maths Chapter 16 Exercise 16.5

RD Sharma Solutions Class 9 Maths Chapter 16 Circles Exercise 16.5 is based on the following topics and subtopics:

- Cyclic Quadrilateral
- Important results:

-The opposite angles of a Cyclic Quadrilateral are supplementary.

– The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.

– A cyclic trapezium is isosceles and its diagonals are equal.

– If two opposite sides of a cyclic quadrilateral are equal, then the other two sides are parallel.